In my differential equations course one of the problems asked the following:

Show that $sin(t^3)$ cannot be a solution of any equation of the form
$$z''' + a_1(t)z'' + a_2(t)z' + a_3(t)z = 0$$
where a1(t), a2(t), and a3(t), are continuous over an open interval containing 0.

The solution was essentially that $z(t) = 0$ was unique by the basic existence and uniqueness theorem, therefore making it so $z=sin(t^3)$ cannot be a solution. However I don't see why the functions could be reversed to argue the opposite.

My professor's explanation was that "$z(t)=0$ is clearly a solution". How can this be answered more rigorously?

$\begingroup$What do you mean by "the functions could be reversed"? In any case $z(t)=0$ is a solution, but if $z(t)=\sin t^3$ then $z(0)=z'(0)=z''(0)=0$ which by the uniqueness theorem is enough to determine $z$.$\endgroup$
– Lord Shark the UnknownFeb 22 '18 at 3:05

1

$\begingroup$Why couldn't I say $z(t) = sin(t^3)$ is a solution and by the uniqueness theorem is unique and therefore $z(t) = 0$ is not a solution?$\endgroup$
– yerpderpingtonFeb 22 '18 at 4:01

$\begingroup$I think Lord Shark's got the idea here. Once you realize that $z$ and its first $3$ derivatives at $0$ are $0$, $z=0$ becomes an obvious solution.$\endgroup$
– MikeFeb 22 '18 at 7:02